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Ques-5-Prove-that-if-a-b-are-any-elements-of-a-group-G-then-the-equation-y-a-b-has-a-unique-solution-in-G-Ques-6-a-Show-that-the-set-G-of-all-non-zero-complex-numbers-i

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Question Number 193759 by Mastermind last updated on 19/Jun/23
Ques. 5 Prove that if a,b are any elements of a group (G, ∗), then the equation y∗a=b has a unique solution in (G, ∗). Ques. 6 a) Show that the set G of all non-zero complex numbers, is a group under multiplication of complex numbers. b) Show that H={a∈G : a_1 ^2 + a_2 ^2 = 1}, where a_1 = Re a and a_2 = Im a is a subgroup of G.
$$\mathrm{Ques}.\:\mathrm{5}\: \\ $$$$\:\:\:\:\:\mathrm{Prove}\:\mathrm{that}\:\mathrm{if}\:\mathrm{a},\mathrm{b}\:\mathrm{are}\:\mathrm{any}\:\mathrm{elements}\:\mathrm{of}\:\:\mathrm{a}\: \\ $$$$\mathrm{group}\:\left(\mathrm{G},\:\ast\right),\:\mathrm{then}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{y}\ast\mathrm{a}=\mathrm{b} \\ $$$$\mathrm{has}\:\mathrm{a}\:\mathrm{unique}\:\mathrm{solution}\:\mathrm{in}\:\left(\mathrm{G},\:\ast\right). \\ $$$$ \\ $$$$\mathrm{Ques}.\:\mathrm{6}\: \\ $$$$\left.\:\:\:\:\:\mathrm{a}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{set}\:\mathrm{G}\:\mathrm{of}\:\mathrm{all}\:\mathrm{non}-\mathrm{zero}\: \\ $$$$\mathrm{complex}\:\mathrm{numbers},\:\mathrm{is}\:\mathrm{a}\:\mathrm{group}\:\mathrm{under} \\ $$$$\mathrm{multiplication}\:\mathrm{of}\:\mathrm{complex}\:\mathrm{numbers}. \\ $$$$ \\ $$$$\left.\:\:\:\:\:\:\mathrm{b}\right)\:\mathrm{Show}\:\mathrm{that}\:\mathrm{H}=\left\{\mathrm{a}\in\mathrm{G}\::\:\mathrm{a}_{\mathrm{1}} ^{\mathrm{2}} \:+\:\mathrm{a}_{\mathrm{2}} ^{\mathrm{2}} \:=\:\mathrm{1}\right\}, \\ $$$$\mathrm{where}\:\mathrm{a}_{\mathrm{1}} \:=\:\mathrm{Re}\:\mathrm{a}\:\mathrm{and}\:\mathrm{a}_{\mathrm{2}} \:=\:\mathrm{Im}\:\mathrm{a}\:\mathrm{is}\:\mathrm{a}\: \\ $$$$\mathrm{subgroup}\:\mathrm{of}\:\mathrm{G}. \\ $$
Answered by aleks041103 last updated on 19/Jun/23
Q.5 (G,∗) is group⇒∀a∈G,∃a^(−1) ∈G: aa^(−1) =a^(−1) a=e ⇒y∗a=b⇒y∗a∗a^(−1) =b∗a^(−1) ⇒y=b∗a^(−1) which is obviously unique. you can proceed also as follows: suppose y_1 ≠y_2 ∈G, s.t. y_1 ∗a=y_2 ∗a=b since ∃a^(−1) ⇒y_1 ∗a∗a^(−1) =y_2 ∗a∗a^(−1) =b∗a^(−1) ⇒y_1 =y_2 ⇒ contradiction ⇒y is unique
$$ \\ $$$${Q}.\mathrm{5} \\ $$$$\left({G},\ast\right)\:{is}\:{group}\Rightarrow\forall{a}\in{G},\exists{a}^{−\mathrm{1}} \in{G}:\:{aa}^{−\mathrm{1}} ={a}^{−\mathrm{1}} {a}={e} \\ $$$$\Rightarrow{y}\ast{a}={b}\Rightarrow{y}\ast{a}\ast{a}^{−\mathrm{1}} ={b}\ast{a}^{−\mathrm{1}} \Rightarrow{y}={b}\ast{a}^{−\mathrm{1}} \\ $$$${which}\:{is}\:{obviously}\:{unique}. \\ $$$${you}\:{can}\:{proceed}\:{also}\:{as}\:{follows}: \\ $$$${suppose}\:{y}_{\mathrm{1}} \neq{y}_{\mathrm{2}} \in{G},\:{s}.{t}. \\ $$$${y}_{\mathrm{1}} \ast{a}={y}_{\mathrm{2}} \ast{a}={b} \\ $$$${since}\:\exists{a}^{−\mathrm{1}} \Rightarrow{y}_{\mathrm{1}} \ast{a}\ast{a}^{−\mathrm{1}} ={y}_{\mathrm{2}} \ast{a}\ast{a}^{−\mathrm{1}} ={b}\ast{a}^{−\mathrm{1}} \\ $$$$\Rightarrow{y}_{\mathrm{1}} ={y}_{\mathrm{2}} \:\Rightarrow\:{contradiction} \\ $$$$\Rightarrow{y}\:{is}\:{unique} \\ $$
Commented by Mastermind last updated on 20/Jun/23
Thank you my boss
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{my}\:\mathrm{boss} \\ $$
Answered by Rajpurohith last updated on 19/Jun/23
Solution:− Let a,b∈G (5)Suppose there are two solutions say y_1 & y_2 of the equation y∗a=b ⇒y_1 ∗a=b & y_2 ∗a=b ⇒y_1 ∗a=y_2 ∗a since a is in a group, a^(−1) exists. ⇒(y_1 ∗a)∗a^(−1) =(y_2 ∗a)∗a^(−1) ⇒By associativity y_1 =y_2 and we are done. ■ (6) (a) say (G, .)=(C−{0}, .) where ′.′ is multiplication of complex numbers which is clearly a binary operation on C. (i)For any z∈C−{0} , z_1 ∈C−{0} be such that z.z_1 =z_1 .z=z we get z_1 =1 which is the identity. (ii)It is obvious that associativity holds. (iii) for each z∈C−{0},z.((1/z))=((1/z)).z=1 hence the inverse of an element z equals ((1/z)) Hence (C−{0}, . ) forms a group. ■ (b) As per given H={z∈C : ∣z∣=1} clearly H⊂C−{0}. We shall use two step subgroup test let z,w∈H . Whether z.w^(−1) ∈H? ∣z.w^(−1) ∣=∣z∣.∣w^(−1) ∣=∣z∣.(1/(∣w∣))=1 (Why?) Hence H is a subgroup of (C−{0}, .} ■
$$\boldsymbol{{Solution}}:−\:{Let}\:{a},{b}\in{G} \\ $$$$\left(\mathrm{5}\right){Suppose}\:{there}\:{are}\:{two}\:{solutions}\:{say}\:{y}_{\mathrm{1}} \:\&\:{y}_{\mathrm{2}} \\ $$$${of}\:{the}\:{equation}\:{y}\ast{a}={b}\: \\ $$$$\Rightarrow{y}_{\mathrm{1}} \ast{a}={b}\:\&\:{y}_{\mathrm{2}} \ast{a}={b} \\ $$$$\Rightarrow{y}_{\mathrm{1}} \ast{a}={y}_{\mathrm{2}} \ast{a}\: \\ $$$${since}\:{a}\:{is}\:{in}\:{a}\:{group},\:{a}^{−\mathrm{1}} \:{exists}. \\ $$$$\Rightarrow\left({y}_{\mathrm{1}} \ast{a}\right)\ast{a}^{−\mathrm{1}} =\left({y}_{\mathrm{2}} \ast{a}\right)\ast{a}^{−\mathrm{1}} \\ $$$$\Rightarrow{By}\:{associativity}\:{y}_{\mathrm{1}} ={y}_{\mathrm{2}} \:{and}\:{we}\:{are}\:{done}.\:\blacksquare \\ $$$$\left(\mathrm{6}\right)\:\left(\boldsymbol{\mathrm{a}}\right)\:{say}\:\left({G},\:.\right)=\left(\mathbb{C}−\left\{\mathrm{0}\right\},\:.\right)\:{where}\:'.'\:{is}\:{multiplication} \\ $$$${of}\:{complex}\:{numbers}\:{which}\:{is}\:{clearly}\:{a}\:{binary}\:{operation}\:{on}\:\mathbb{C}. \\ $$$$\left({i}\right){For}\:{any}\:{z}\in\mathbb{C}−\left\{\mathrm{0}\right\}\:,\:{z}_{\mathrm{1}} \in\mathbb{C}−\left\{\mathrm{0}\right\}\:\:{be}\:{such}\:{that}\:{z}.{z}_{\mathrm{1}} ={z}_{\mathrm{1}} .{z}={z} \\ $$$${we}\:{get}\:{z}_{\mathrm{1}} =\mathrm{1}\:{which}\:{is}\:{the}\:{identity}. \\ $$$$\left({ii}\right){It}\:{is}\:{obvious}\:{that}\:{associativity}\:{holds}. \\ $$$$\left({iii}\right)\:{for}\:{each}\:{z}\in\mathbb{C}−\left\{\mathrm{0}\right\},{z}.\left(\frac{\mathrm{1}}{{z}}\right)=\left(\frac{\mathrm{1}}{{z}}\right).{z}=\mathrm{1} \\ $$$${hence}\:{the}\:{inverse}\:{of}\:{an}\:{element}\:{z}\:{equals}\:\left(\frac{\mathrm{1}}{{z}}\right) \\ $$$${Hence}\:\left(\mathbb{C}−\left\{\mathrm{0}\right\},\:.\:\right)\:{forms}\:{a}\:{group}.\:\blacksquare \\ $$$$\left(\boldsymbol{\mathrm{b}}\right)\:{As}\:{per}\:{given}\:\:{H}=\left\{{z}\in\mathbb{C}\::\:\mid{z}\mid=\mathrm{1}\right\} \\ $$$${clearly}\:\:{H}\subset\mathbb{C}−\left\{\mathrm{0}\right\}. \\ $$$${We}\:{shall}\:{use}\:{two}\:{step}\:{subgroup}\:{test} \\ $$$${let}\:{z},{w}\in{H}\:\:.\:\:{Whether}\:{z}.{w}^{−\mathrm{1}} \in{H}? \\ $$$$\:\mid{z}.{w}^{−\mathrm{1}} \mid=\mid{z}\mid.\mid{w}^{−\mathrm{1}} \mid=\mid{z}\mid.\frac{\mathrm{1}}{\mid{w}\mid}=\mathrm{1}\:\left({Why}?\right) \\ $$$${Hence}\:{H}\:{is}\:{a}\:{subgroup}\:{of}\:\left(\mathbb{C}−\left\{\mathrm{0}\right\},\:.\right\}\:\blacksquare \\ $$$$ \\ $$
Commented by Mastermind last updated on 20/Jun/23
I do really appreciate it
$$\mathrm{I}\:\mathrm{do}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{it} \\ $$
Commented by Rajpurohith last updated on 21/Jun/23
Please do post Group theory questions, I′m taking a course again in Abstract algebra, It helps me too.
$${Please}\:{do}\:{post}\:{Group}\:{theory}\:{questions}, \\ $$$${I}'{m}\:{taking}\:{a}\:{course}\:{again}\:{in}\:{Abstract}\:{algebra}, \\ $$$${It}\:{helps}\:{me}\:{too}. \\ $$
Commented by Mastermind last updated on 22/Jun/23
Wow! that′s good, i′ll be posting it
$$ \\ $$$$\mathrm{Wow}! \\ $$$$\mathrm{that}'\mathrm{s}\:\mathrm{good},\:\mathrm{i}'\mathrm{ll}\:\mathrm{be}\:\mathrm{posting}\:\mathrm{it} \\ $$

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